# How do you solve sin(x/2)-1=0 between the interval 0<=x<8pi?

$x = \pi$ or $x = 5 \pi$
Since $0 \le x < 8 \pi$, we have $0 \le \frac{x}{2} < 4 \pi$. Since $\sin \left(\frac{x}{2}\right) = 1$, the only allowed values of $\frac{x}{2}$ in the interval $\left[0 , 4 \pi\right)$ are $\frac{x}{2} = \frac{\pi}{2}$ and $\frac{x}{2} = \frac{5 \pi}{2}$.